Applied aeronautics; the airplane - Beeldbibliotheek

:—12 APPLIED AERONAUTICSor a velocity of a little over 55 m.p.li. Increasing his angle of attackstill further, or at about 8 (leg., which is the lowest point on the curve,where the horsepower required for horizontal flight is only 30, we geta point of most economical flight. Tlien. as we decrease the angle ofattack, the power required rises rapidly until at 40 m.p.h. the two curvescross again and any increase in the angle of attack would cause the machineto stall in the sense of going down, which might take the form ofeither a nose dive or tail slip. It is well to compare this with Fig. 20.F'kj. 24Calculation of power required to rVunbIt is also interesting to compare this with Fig. 23, taken fromLangley's The Stored Euergi/ of the Wind, and which illustrates therapid changes in the velocity of the wind occurring in short intervals oftime. The vertical lines represent spaces of one minute and the horizontallines wind speeds differing by 4 m.p.h. It will be noticed that between32 and 24 min. the wind fell from about 37 m.p.h. to 12 m.i).h. and roseagain to 38 m.p.h. On account of the momentum of the airplane it wouldbe practically impossible for its actual velocity to change with anythinglike that rapidity, and as the lift depends upon the square of the velocityit is evident that the pilot would experience a series of ''bumps" whenthe velocity increased, and momentary drops when the velocity decreased.The feeling has been likened to a motor boat driving rapidly through achoppy sea.Power to ClimbSuppose the center of gravity of a machine be moving in the directionAB, Fig. 24, with a velocity of V miles per hour. The horsepowerwill then be the sum of two components, viz., that necessary to overcomethe wind resistance, as already given for horizontal flight and that necessaryto lift the machine through the distance CB in the time required forthe machine to travel fiom A to B.Now if AB be taken to represent thedistance the machine travels in an liour, BC would then represent thevelocity of climb. The power consumed in climbing is equal to theproduct of the weight of the machine in pounds by the velocity of climbin miles per hour divided by 375. Let us call AB/BC the climbing ratioR which gives us BC = AB/ K = V/R. We will have then the i>owerexpended in the climb alone equal to WV/R, and the total horsepowerbecomes